W=Fd. Force is not given so we solve for it. F=ma, m=5kg, a=2m/s^2, F=10N. Distance is not given so we solve for it, x=.5a(t^2)=.5(2)(7x7)=49m. F=10N, d=49m, W=490J.
Answer : The process of changing a property of a wave to transmit information is called Modulation.
Explanation :
Modulation is the process of changing the property of wave to transmit information. This is done with the help of modulator.
In modulation, the message signal is superimposed on a high frequency signal. A sine wave ( usually high frequency ) is used as a high frequency carrier wave.
Modulation can be done in many ways like :
(1) Frequency modulation
(2) Amplitude modulation
(3) Pulse modulation
The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.
For this "type" of motion, displacement (Δx) can be determined by:
is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
= - 42
Displacement of the boat for t=6.0s interval is = - 42m, i.e., 42 m to the left.
Electric field strength = resistivity of copper x current density
where
p= 1.72 x 10^-8 <span>ohm meter
diameter = 2.05mm=.00205 m
current = 2.75 A
</span>get first the current density:
current density = current/ cross section area
find the cross section area
cross section area = pi.(d/2)^2;
cross section = 3.3 006x10-6 m^2
substitute the values
current density = 2.75A/3.3006x 10-6m^2
current density=35.55 x1 0^2 A/m^2
Electric field stregnth =1.72 x 10^-8 ohm meter x 35.55 x10^2 A/m^2
Electric field stregnth= 46.415 Volts/m
The electric field strength of copper is 46.415 V/m.
Question:
Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. In which direction should an arrow at position 1 point?
left
right
up
down
Answer:
The correct answer is
Left
Explanation:
Magnetic circuits describe the path of a magnetic flux. In the same way electricity follows a complete closed circuit, the path of a magnetic flux is also a complete and closed circuit which leaves from the N pole, migrates through the air and reenters the magnet through the S pole through which it passes back into the magnet to come to the N pole again.
As such the magnetic field lines emanate from the N pole which is on he right to the S pole which is on the left. Hence the arrow should point in the left direction.
